n(n-1)(n+1)=1320

Solution for n(n-1)(n+1)=1320 equation:

n(n-1)(n+1)=1320

We move all terms to the left:

n(n-1)(n+1)-(1320)=0

We use the square of the difference formula

n^2-1-1320=0

We add all the numbers together, and all the variables

n^2-1321=0

a = 1; b = 0; c = -1321;
Δ = b2-4ac
Δ = 02-4·1·(-1321)
Δ = 5284
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$n_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$n_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

The end solution:

$\sqrt{\Delta}=\sqrt{5284}=\sqrt{4*1321}=\sqrt{4}*\sqrt{1321}=2\sqrt{1321}$
$n_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(0)-2\sqrt{1321}}{2*1}=\frac{0-2\sqrt{1321}}{2}
=-\frac{2\sqrt{1321}}{2}
=-\sqrt{1321}
$
$n_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(0)+2\sqrt{1321}}{2*1}=\frac{0+2\sqrt{1321}}{2}
=\frac{2\sqrt{1321}}{2}
=\sqrt{1321}
$